F(x)= ln ((3t + 1)^4 / (2t + 1)^3)

F'(x)=

To find the derivative of the function F(x), which is defined as F(x) = ln((3t + 1)^4 / (2t + 1)^3), we can use the differentiation rules and properties of logarithmic functions. Here's how you can get the answer:

Step 1: Simplify the function
Start by simplifying the function using logarithmic properties. Recall that ln(a/b) = ln(a) - ln(b). Applying this property to the function, we have:

F(x) = 4ln(3t + 1) - 3ln(2t + 1)

Step 2: Apply the differentiation rules
To differentiate F(x), we can apply the chain rule for composite functions and the derivative of ln(x).

The derivative of ln(u) with respect to t is given by d/dt[ln(u)] = (1/u) * du/dt.

So, let's differentiate each term individually:

For the first term, 4ln(3t + 1):
Using the chain rule, we have:
d/dt[4ln(3t + 1)] = 4 * d/dt[ln(3t + 1)] = 4 * (1 / (3t + 1)) * d/dt[3t + 1]

For the second term, -3ln(2t + 1):
Similarly, using the chain rule, we have:
d/dt[-3ln(2t + 1)] = -3 * d/dt[ln(2t + 1)] = -3 * (1 / (2t + 1)) * d/dt[2t + 1]

Step 3: Evaluate the derivatives
Now, let's find the derivatives of the expressions inside the derivatives from Step 2:

For the first term, d/dt[3t + 1]:
The derivative of 3t + 1 with respect to t is simply 3.

For the second term, d/dt[2t + 1]:
The derivative of 2t + 1 with respect to t is 2.

Step 4: Substitute the derivatives back into the expressions
Putting everything together, we have:

F'(x) = 4 * (1 / (3t + 1)) * 3 - 3 * (1 / (2t + 1)) * 2

Simplifying further, we get:

F'(x) = (12 / (3t + 1)) - (6 / (2t + 1))

Therefore, the derivative of F(x) with respect to t is F'(x) = (12 / (3t + 1)) - (6 / (2t + 1)).